Let $G=\langle x\rangle$ be cyclic of order $n$. Prove that $\langle x^r\rangle⊆\langle x^s\rangle$ iff r is a multiple of $s$ modulo $n$.

Question

Let $G=\langle x\rangle$ be cyclic of order $n.$ Prove that $\langle x^r\rangle\subseteq \langle x^s\rangle \iff r$ is a multiple of $s$ modulo $n.$

I know you have to approach this from both ways, $\Longrightarrow$ and $\Longleftarrow$ but im not sure of the correlation from both ways, can someone help me?

Answer 1

1. If $\subseteq$, then $x^r\in \langle x^s\rangle=\{x^{sk}:k\in\mathbb{N}\}$. Hence there is some $k\in\mathbb{N}$ such that $x^r=x^{sk}$. Since a group is cancellative, we have $1=x^{sk-r}$. Since $G$ is cyclic of order $n$, then $n|sk-r$ or $sk\equiv r\pmod{n}$.

2. The other direction is simple.